I am trying to reduce and obtain a simple expression from following sum
$$ \sum_{i = 1}^{\infty}{1 \over i}\, {\,\mathrm{e}^{-\lambda}\,\lambda^{i} \over i!}\, {1 \over 1 - \,\mathrm{e}^{-\lambda}\,} $$
Problem can be put in general form as:
$$ \sum_{i = c}^{\infty}{1 \over i}\, {\,\mathrm{e}^{-\lambda}\,\lambda^{i} \over i!\left(\displaystyle{\, 1 - \sum_{k = 1}^{c - 1} {\,\mathrm{e}^{-\lambda}\,\lambda^{k} \over k!}\,}\right)}\,, \qquad c \geq 1 $$
The first sum is $$ e^{-\lambda}\int_0^\lambda\frac{e^x-1}{x}\,\mathrm{d}x $$ which is not an elementary function. However, we can use the Exponential Integral to get $$ e^{-\lambda}\sum_{k=1}^\infty\frac{\lambda^k}{k\,k!}=e^{-\lambda}\left(\mathrm{Ei}(\lambda)-\gamma-\log\left|\lambda\right|\right) $$